Question

determine the limiting reactant for the reaction. mg(s) o2(g) mgo(s) previous answers correct the limiting reactant is the reactant that can form the least amount of product. the given mass of mg will form less mgo than the given mass of o2. part b determine the theoretical yield for the reaction. express your answer in grams. m = g

Explanation:

Step1: Write the balanced chemical equation

The reaction between magnesium ($Mg$) and oxygen ($O_2$) to form magnesium - oxide ($MgO$) is $2Mg(s)+O_2(g)
ightarrow2MgO(s)$.

Step2: Calculate the moles of reactants

Let's assume we have masses of $Mg$ and $O_2$ given. If the mass of $Mg$ is $m_{Mg}$ and its molar - mass $M_{Mg}=24.31\ g/mol$, then the number of moles of $Mg$, $n_{Mg}=\frac{m_{Mg}}{24.31\ g/mol}$. If the mass of $O_2$ is $m_{O_2}$ and its molar - mass $M_{O_2} = 32.00\ g/mol$, then the number of moles of $O_2$, $n_{O_2}=\frac{m_{O_2}}{32.00\ g/mol}$.

Step3: Determine the moles of product formed from each reactant

From the balanced equation, 2 moles of $Mg$ produce 2 moles of $MgO$, so the moles of $MgO$ formed from $Mg$, $n_{MgO - from - Mg}=n_{Mg}$. And 1 mole of $O_2$ produces 2 moles of $MgO$, so the moles of $MgO$ formed from $O_2$, $n_{MgO - from - O_2}=2n_{O_2}$.

Step4: Calculate the mass of the product

The molar - mass of $MgO$ is $M_{MgO}=24.31\ g/mol + 16.00\ g/mol=40.31\ g/mol$. The mass of $MgO$ formed, $m_{MgO}=n_{limiting - reactant}\times M_{MgO}$, where $n_{limiting - reactant}$ is the number of moles of the limiting reactant (in this case, $Mg$). If $n_{Mg}$ is the limiting reactant, then $m_{MgO}=n_{Mg}\times40.31\ g/mol$.

Since we know $Mg$ is the limiting reactant from part A, assume we have $n_{Mg}$ moles of $Mg$. Then the theoretical yield of $MgO$ is $m = n_{Mg}\times40.31$. But since no masses of reactants are given in the problem, we'll express the answer in terms of the moles of $Mg$. If the number of moles of $Mg$ is $n_{Mg}$, the theoretical yield of $MgO$ is $m = 40.31n_{Mg}\ g$.

Answer:

We need the mass of $Mg$ (or number of moles of $Mg$) to give a numerical answer. If the number of moles of $Mg$ is $n_{Mg}$, the theoretical yield $m = 40.31n_{Mg}\ g$.